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Dependent on what side you are given you would use Sin(Θ) = Opposite/Hypotenuse just rearrange the formula to Hypotenuse = Opposite/Sin(Θ). Or if you are given the adjacent side use Cosine(Θ)=Adjacent/Hypotenuse, then: Hypotenuse = Adjacent/Cosine(Θ)
The basic functions are sine, cosine, tangent, cosecant, secant and cotangent. In addition, there are their inverses, whose full names use the prefix "arc" [arcsine, arc cosine, etc] but are more often written as sin-1, cos-1 and so on.
Sin(20) = You need either a scientific calculator or Castles four figure Tables. Using a scientific calculator Sin ( 20 ) = 0.342020143.... Sin(20) ~ 0.3420 (4d.p.).
There are two ways to solve for the double angle formulas in trigonometry. The first is to use the angle addition formulas for sine and cosine. * sin(a + b) = sin(a)cos(b) + cos(a)sin(b) * cos(a + b) = cos(a)cos(b) - sin(a)sin(b) if a = b, then * sin(2a) = sin(a)cos(a) + cos(a)sin(a) = 2sin(a)cos(a) * cos(2a) = cos2(a) - sin2(b) The cooler way to solve for the double angle formulas is to use Euler's identity. eix = cos(x) + i*sin(x). Yes, that is "i" as in imaginary number. we we put 2x in for x, we get * e2ix = cos(2x) + i*sin(2x) This is the same as * (eix)2 = cos(2x) + i*sin(2x) We can substitute our original equation back in for eix. * (cos(x) + i*sin(x))2 = cos(2x) + i*sin(2x) We can distribute the squared term. * cos2(x) + i*sin(x)cos(x) + i*sin(x)cos(x) + (i*sin(x))2 = cos(2x) + i*sin(2x) And simplify. Because i is SQRT(-1), the i squared term becomes negative. * cos2(x) + 2i*sin(x)cos(x) - sin2(x) = cos(2x) + i*sin(2x) * cos2(x) - sin2(x) + 2i*sin(x)cos(x) = cos(2x) + i*sin(2x) Now you can plainly see both formulas in the equation arranged quite nicely. I don't yet know how to get rid of the i, but I'm working on it.
It depends on what else you know. If it is a non-right triangle, and you only know angle a, it is impossible to fing side A (the side opposite an angle usually has the same letter, but capitalized). If you know the other two sides, then I would use the law of cosines: For a triangle with sides A B C A = √(B2+C2-(2*B*C*(cos (a)))) If you know another angle and one side, I would use the law sines: A/(sin a) = B/(sin b) therefore, A = (sin a) * B/(sin b) If it is a right triangle, and you know another side, than your job is even easier: If you know the hypotenuse (side C), than: A = C *(sin a) If you know the adjacent side (side B), than: A = B * (tan a)